# Sum of elements in a sequence

Let $a_n$ be a sequence in $\mathbb{R}$ and $a\in\mathbb{R}$. Suppose that $N \in \mathbb{N}$, $\epsilon >0$ and for every $n > N$ $|a_n -a|<\epsilon$. Show that for every $n>N$ the following holds: $$\left|\frac{1}{n} \sum_{k=N+1}^n a_k -a \right|\leq \epsilon +\frac{N}{n} |a|$$

I have the following but that would lead to an absolute smaller and the expression doesn't give the right answer. I wonder where I've gone wrong.

\begin{align*} \left|\frac{1}{n} \sum_{k=N+1}^n a_k -a \right| &= \left| \frac{1}{n} \sum_{k=N+1}^n (a_k -a +a) -a \right|\\\\ & \leq \left|\frac{1}{n} \sum_{k=N+1}^n (|a_k - a| +|a|) -a \right|\\\\ & < \left|\frac{1}{n} \sum_{k=N+1}^n (\epsilon+|a|) -a \right| \\\\ &= \left|\frac{1}{n} (n -(N+1))(\epsilon +a) -a \right|\\\\ & = \left|\frac{n-N-1}{n}(\epsilon +a) -a\right| \end{align*}

• $\sum_{k=N+1}^n(\epsilon + |a|) = (n - N)(\epsilon + |a|)$, rather than $(n - N - 1)(\epsilon + |a|)$. – PSPACEhard May 30 '15 at 10:50
• But then still I get a slightly different expression than what was asked for? is it still correct or have I made another mistake? I understand that if something is $<$ then it's also $\leq$ but I suppose that the question would then state $<$ and not $\leq$. – DeanTheMachine May 31 '15 at 9:27
• I think only $<$ holds. Let me prove this by contradiction. Assume that $$\left| \frac{1}{n}\sum_{k=N+1}^n a_k - a\right| = \epsilon + \frac{N}{n}|a|$$. For the left side, we have $$\left| \frac{1}{n}\sum_{k=N+1}^n a_k - a\right| \leq \frac{1}{n} (n - N) \epsilon = (1 - \frac{N}{n})\epsilon$$. This implies that $$\epsilon + \frac{N}{n}|a| \leq (1 - \frac{N}{n})\epsilon$$, which can't be true. Q.E.D. – PSPACEhard May 31 '15 at 10:44

$\quad \left|\frac{1}{n} \sum_{k=N+1}^n a_k -a \right|\\ = \left| \frac{1}{n} \sum_{k=N+1}^n (a_k - a) -a + \frac{1}{n}\sum_{k=N+1}^na\right|\\ \leq \frac{1}{n}\sum_{k=N+1}^n\left|a_k - a\right| +\left|\frac{n - N}{n}a - a\right|\\ < \frac{1}{n} (n - N)\epsilon + \left|-\frac{N}{n}a\right|\\ = \epsilon - \frac{N}{n}\epsilon + \frac{N}{n}\left|a\right|\\ \leq \epsilon + \frac{N}{n}|a|$